# Prove the following claim about skew symmetric matrix $\langle x,Ax\rangle =0$

The real $$n\times n$$ matrix $${\textstyle A}$$ is skew-symmetric if and only if

$${\displaystyle \langle Ax,y\rangle =-\langle x,Ay\rangle \quad \forall x,y\in \mathbb {R} ^{n}}$$ This is also equivalent to $${\textstyle \langle x,Ax\rangle =0}$$ for all $${\displaystyle x\in \mathbb {R} ^{n}}$$

I am trying to show $$\langle x,Ax\rangle =0$$ but I don't really see it. If I had something like $$\langle x,Ax\rangle = -\langle x,A x\rangle$$ then I can claim $$\langle x,Ax\rangle = 0$$, but instead I have $$\langle x,Ax\rangle = -\langle x,A^\top x\rangle = -\langle Ax, x\rangle \neq 0 \quad (?)$$

Note that $$\langle x,Ax\rangle\stackrel{\text{skew symmetry}}{=}-\langle Ax,x\rangle= -\langle x,Ax\rangle\\ \implies 2\langle x,Ax\rangle=0\\ \implies \langle x,Ax\rangle=0$$